Question 1 |
While minimizing the function f(x), necessary and sufficient conditions for a point, x_{0} to be a minima are:
{f}'(x_{0}) \gt 0\: and\: \: {f}''(x_{0})=0 | |
{f}'(x_{0}) \lt 0\: and\: \: {f}''(x_{0})=0 | |
{f}'(x_{0})=0\: and\: \: {f}''(x_{0}) \lt 0 | |
{f}'(x_{0})=0\: and\: \: {f}''(x_{0}) \gt 0 |
Question 1 Explanation:
f(x) has a local minimum at x=x_{0}
if {f}'\left ( x_{0} \right )=0
and {f}''\left ( x_{0} \right ) \gt 0
if {f}'\left ( x_{0} \right )=0
and {f}''\left ( x_{0} \right ) \gt 0
Question 2 |
In Newton-Raphson iterative method, the initial guess value ( x_{ini}) is considered as zero while finding the roots of the euation:f(x)=-2+6x-4x^{2}+0.5x^{3}. The correction, \Delta x, to be added to x_{ini} in the first iteration is____________.
0.5 | |
0.33 | |
0.8 | |
0.2 |
Question 2 Explanation:
\begin{aligned} f\left ( x \right )&=-2+6x-4x^{2}+0.5x^{3} \\ {f}'\left ( x \right )&=6-8x+1.5x^{2} \\ x_{ini}&=0\\ \text{By Newton }&\text{Raphson Method,}\\ x_{1}&=x_{ini}-\frac{f\left ( x_{ini} \right )}{{f}'\left ( x_{ini} \right )} \\ &= 0-\frac{-2}{6} \\ \Rightarrow \;\;x_{1}&=\frac{1}{3} \\ \therefore \;\; \Delta x&=x_{1}-x_{ini}=\frac{1}{3} \end{aligned}
Question 3 |
Given , i=\sqrt{-1}, the value of the definite integral, I=\int_{0}^{\pi/2}\frac{\cos x+i\sin x}{\cos x-i\sin x}dx is:
1 | |
-1 | |
i | |
-i |
Question 3 Explanation:
\begin{aligned} I&=\int_{0}^{\pi /2}\frac{\cos x+i\sin x}{\cos x-i\sin x}dx \\ &=\int_{0}^{\pi /2}\frac{e^{ix}}{e^{-ix}}dx=e^{2ix}dx \\ &=\frac{1}{2i}\left [ e^{2ix} \right ]_{0}^{\pi /2} \\ &=\frac{1}{2i}\left [ e^{i\pi }-1 \right ] \\ &=\frac{1}{2i}\left ( -1-1 \right ) \\ &=-\frac{1}{i}=i \end{aligned}
Question 4 |
\lim_{x\rightarrow \infty }\left ( 1+\frac{1}{x} \right )^{2x} is equal to
e^{-2} | |
e | |
1 | |
e^{2} |
Question 4 Explanation:
y=\: \lim_{x\rightarrow \infty }\left ( 1+\frac{1}{x} \right )^{2x}
\log y = \lim_{x\rightarrow \infty }2x\log \left ( 1+\frac{1}{x} \right )
Which is in the form of \infty \times 0.
To convert this into \frac{0}{0} form, we rewrite as,
\Rightarrow \log y= \lim_{x\rightarrow \infty }\frac{2\log \left ( 1+\frac{1}{x} \right )}{1/x}
Now is in \frac{0}{0} form.
Using L'Hospital's Rule,
\begin{aligned} \log y&=\lim_{x\rightarrow \infty }\frac{\frac{2\times -\frac{1}{x^{2}}}{1+\frac{1}{x}}}{-\frac{1}{x^{2}}} \\ \log y&=\lim_{x\rightarrow \infty }\frac{2}{1+\frac{1}{x}}=2 \\ \therefore \;\; y&= e^{2}\end{aligned}
\log y = \lim_{x\rightarrow \infty }2x\log \left ( 1+\frac{1}{x} \right )
Which is in the form of \infty \times 0.
To convert this into \frac{0}{0} form, we rewrite as,
\Rightarrow \log y= \lim_{x\rightarrow \infty }\frac{2\log \left ( 1+\frac{1}{x} \right )}{1/x}
Now is in \frac{0}{0} form.
Using L'Hospital's Rule,
\begin{aligned} \log y&=\lim_{x\rightarrow \infty }\frac{\frac{2\times -\frac{1}{x^{2}}}{1+\frac{1}{x}}}{-\frac{1}{x^{2}}} \\ \log y&=\lim_{x\rightarrow \infty }\frac{2}{1+\frac{1}{x}}=2 \\ \therefore \;\; y&= e^{2}\end{aligned}
Question 5 |
Let \mathbf{A}=\left [ a_{ij} \right ],\; \; \; 1\leq i, \; j\leq n with n\geq 3\; and\; a_{ij}=i.j. The rank of A is:
0 | |
1 | |
n-1 | |
n |
Question 5 Explanation:
Rank of A=1
Because each row will be scalar multiple of first row.So we will get only one-zero row in row Echeleaon form of A.
Alternative:
Rank of A=1
Because all the minors of order greater than 1 will be zero.
Because each row will be scalar multiple of first row.So we will get only one-zero row in row Echeleaon form of A.
Alternative:
Rank of A=1
Because all the minors of order greater than 1 will be zero.
There are 5 questions to complete.